**NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions**

In this **exercise 3.4 of the trigonometric function,** we will study the solutions of **trigonometric equations**. The equation containing **trigonometric functions** are known as **trigonometric equations,**like sinx = siny, cosx +1 =0 etc. The solutions of these **trigonometric equations** are of two kinds.

**1-Principal Solution.** The values of an unknown angle between 0° and 360°that satisfy the given **equation** are known as **principal solutions** of the** given equation.**

**2-General Solutions.** Set of infinite **solutions** that satisfy the given **equation** is known as the **general** **solution,** this **solution** is lead due to the periodicity of the **trigonometric functions**. The **general solution** depends upon period of the **trigonometric function**, as an example if **principal solution** of the **equation** sinx =1 ⇒ sinx= sinπ/2,so **principal solution** is x =π/2 then, this **equation** will satisfy for x= (2π ±π/2), (4π± π/2)…..etc, so in compact form, we can write these **solutions** x =nπ +(–1)n×π/2, it is all about the **general solution**. The periodicity and **general solutions of trigonometric functions** are given below.

**CBSE Class 11-Question paper of maths 2015**

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**NCERT Solutions for Class 11 Maths Exercise 3.4 of Chapter 3 Trigonometric Functions**

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**Q1. Find the principal and general solution of the equation tanx = √3**.

**Answer.**

**Hint: All possible values of x lying between 0 and 2π which satisfy the equation are called principal solutions.**

**Therefore, substituting √3 by tanπ/3 and tan 4π/3 we get**

**For calculating general solution if we have tanx = tany then the value of x = nπ + y is known as general solution of the equation tanx = tany,so**

**Taking the smaller value of the solution π/3**

**4π/3 was not taken here because it is one of the infinite solutions ( for n=1, x = π +π/3 =4π/3)**

**(n∈ z)**

**will be the general solution of the equation.**

**Q2. Find the principal and general solution of the equation secx = 2.**

**Answer.**

**secx =2**

**Since the value of secx is positive(i.e 2),x will lie on 1 st and 4 th quadrant.(value of secx is positive in 1st and 4 th quadrant).**

**4 th quadrant means 2π −x (x =π/3)**

**Therefore principal solutions of the equation will be**

**If secx = secy then general solution of the equation is given by the same expression as for the cosx =cosy, see the hint.**

**Hint: If cosx = cosy⇒1/secx = 1 /secy⇒secx=secy,therefore secx =secy implies that cosx = cosy.**

**(n ∈ z)**

**Therefore the general solution of the equation will be 2nπ ± π/3.**

**Q3. Find the principal and general solutions of the equation cotx = –√3.**

**Answer.**

**cotx = –√3.**

**√3 = cotπ/6**

**Value of cotx is negative ,so x will lie on 2nd quadrant (i.e π– π/6) and fourth quadrant(i.e 2π– π/6)(value of cotx is negative in 2 nd and 4 th quadrant).**

**Therefore the principal solutions of the given equation will be,**

**Therefore the general solution of the given equation is**

**Q4.Find the general solution of cosecx =–2.**

**Answer.**

**Value of cosec is negative in 3rd(π+π/6) and 4 th quadrant (2π–π/6).**

**Therefore the principal solutions of the given equation will be,**

**The general solution of the given equation will be**

**As we know cosecx = 1/sinx**

**Q5.Find the general solution of the equation cos4x = cos2x.**

**Answer.**

**cos4x = cos2x**

**cos4x – cos2x = 0**

**⇒2sin3x.sinx= 0**

**⇒sin3x=0 and sinx =0**

**sin3x=0**

**sin3x = sin0**

**since θ = 0**

**3x = nπ**

**sinx =0 =sin0**

**x = nπ**

**Therefore general solution of both the equation will be**

**Q6. Find the general solution of the equation cos3x + cosx – cos2x = 0.**

**Answer.**

**cos3x + cosx – cos2x = 0**

**(cos3x + cosx )– cos2x = 0**

**⇒2cos2x.cosx – cos2x =0**

**⇒cos2x(2cosx – 1) = 0**

**Since general solution of the equation cosx = cosy is x = 2nπ ± y,so**

**cos2x = 0**

**2cosx – 1 =0**

**2cosx = 1**

**Therefore the general solution of the given equation will be**

**Q7. Find the general solution of the equation sin2x + cosx = 0.**

**Answer.**

**sin2x + cosx = 0**

**2sinx.cosx + cosx = 0**

**cosx(2sinx + 1) = 0**

**cosx = 0, 2sinx + 1 = 0**

**cosx = cos0, sinx = –1/2⇒sinx = -sinπ/6 = sin(π+ π/6)=sin7π/6**

**Therefore the general solution of the given equation will be ,**

**Q8. Find the general solution of the equation sec²2x = 1 – tan2x.**

**Answer.**

**sec²2x = 1 – tan2x**

**1 + tan²2x = 1 – tan2x**

**tan²2x + tan2x = 0**

**tan2x(tan2x +1) = 0**

**tan2x = 0**

**tan2x = tan0( general solution of the equation (tanx = tany) is x = nπ + y)**

**2x = nπ + 0**

**tan2x +1 = 0**

**tan2x = –1**

**Value of tan is negative so 2x will lye in 2nd(π –π/4) and 4 th quadrant (2π – π/4)[tanθ is negative in 2nd and 4 th quadrant].**

**Taking the smaller value of 2x**

**Therefore the general solution of the given equation is**

(**n εz)**

**Q9. Find the general solution of the equation sinx + sin3x + sin5x = 0.**

**Answer.**

**sinx + sin3x + sin5x = 0**

**(sinx + sin5x )+ sin3x = 0**

**⇒2sin3x.cos(–4x) + sin3x=0**

**⇒2sin3x.cos4x +sin3x=0**

**⇒sin3x(2cos4x +1) =0**

**sin3x = 0 and 2cos4x +1 =0⇒cos4x = –1/2=–cosπ/3**

**cos4x = –cosπ/3**

**Value of cos4x is negative in 2nd (π -π/3) and 3rd quadrant(π + π/3),taking smaller value in getting general solution.**

**Therefore the general solution of the given equation is**

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**NCERT Solutions for class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

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**NCERT solutions for class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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